Estimate the following limit using graphs or tables. In (1 + h) lim h h?0 In (1 + h) lim h h- (Type an integer or decimal rounded to the nearest hundredth as needed.)
The Correct Answer and Explanation is:
Sure! Your question seems to be about estimating the limit of the expression: limh→0ln(1+h)h\lim_{h \to 0} \frac{\ln(1 + h)}{h}
This is a classic limit related to the derivative of the natural logarithm function.
Step 1: Understand the expression
The expression is: ln(1+h)h\frac{\ln(1 + h)}{h}
and we want to find limh→0ln(1+h)h\lim_{h \to 0} \frac{\ln(1 + h)}{h}
As h→0h \to 0, the expression takes the indeterminate form 00\frac{0}{0}, since ln(1+0)=ln(1)=0\ln(1 + 0) = \ln(1) = 0 and the denominator is zero as well.
Step 2: What does the limit represent?
This limit is actually the definition of the derivative of ln(x)\ln(x) at x=1x = 1: ddxlnx∣x=1=limh→0ln(1+h)−ln(1)h=limh→0ln(1+h)h\frac{d}{dx} \ln x \bigg|_{x=1} = \lim_{h \to 0} \frac{\ln(1 + h) – \ln(1)}{h} = \lim_{h \to 0} \frac{\ln(1 + h)}{h}
Since ln(1)=0\ln(1) = 0.
We know from calculus that ddxlnx=1x\frac{d}{dx} \ln x = \frac{1}{x}
Therefore, ddxlnx∣x=1=1\frac{d}{dx} \ln x \bigg|_{x=1} = 1
So, the limit should be 1.
Step 3: Estimate the limit using a table
Let’s verify this by plugging in small values of hh close to zero (both positive and negative) and calculate the value of ln(1+h)h\frac{\ln(1 + h)}{h}:
| hh | ln(1+h)\ln(1+h) | ln(1+h)h\frac{\ln(1+h)}{h} |
|---|---|---|
| 0.1 | ln(1.1)≈0.0953\ln(1.1) \approx 0.0953 | 0.0953/0.1=0.9530.0953 / 0.1 = 0.953 |
| 0.01 | ln(1.01)≈0.00995\ln(1.01) \approx 0.00995 | 0.00995/0.01=0.9950.00995 / 0.01 = 0.995 |
| 0.001 | ln(1.001)≈0.0009995\ln(1.001) \approx 0.0009995 | 0.0009995/0.001=0.99950.0009995 / 0.001 = 0.9995 |
| -0.1 | ln(0.9)≈−0.10536\ln(0.9) \approx -0.10536 | −0.10536/−0.1=1.0536-0.10536 / -0.1 = 1.0536 |
| -0.01 | ln(0.99)≈−0.01005\ln(0.99) \approx -0.01005 | −0.01005/−0.01=1.005-0.01005 / -0.01 = 1.005 |
| -0.001 | ln(0.999)≈−0.0010005\ln(0.999) \approx -0.0010005 | −0.0010005/−0.001=1.0005-0.0010005 / -0.001 = 1.0005 |
Step 4: Interpretation
As h→0h \to 0, from both positive and negative sides, the ratio ln(1+h)h\frac{\ln(1+h)}{h} approaches 1. The values get closer and closer to 1, confirming the limit.
Final answer:
1.00\boxed{1.00}
Explanation (300 words):
The limit limh→0ln(1+h)h\lim_{h \to 0} \frac{\ln(1+h)}{h} is a foundational limit in calculus that connects the natural logarithm function and its derivative. This limit represents the derivative of the natural logarithm function at the point x=1x = 1. The natural logarithm function ln(x)\ln(x) is defined for positive xx, and its derivative is known to be 1/x1/x.
To understand this limit, notice that as hh approaches zero, both the numerator ln(1+h)\ln(1+h) and denominator hh approach zero, giving an indeterminate form 0/00/0. This situation invites the use of methods like L’Hôpital’s Rule or the definition of the derivative itself.
If we interpret the limit as a derivative definition, it is: limh→0ln(1+h)−ln(1)h\lim_{h \to 0} \frac{\ln(1+h) – \ln(1)}{h}
which is the slope of the tangent line to the curve y=ln(x)y = \ln(x) at x=1x=1. Since ln(1)=0\ln(1) = 0, the expression simplifies nicely.
Knowing that the derivative of ln(x)\ln(x) is 1/x1/x, at x=1x=1 the derivative equals 1. This matches the numerical values we get when plugging in very small values of hh in the expression ln(1+h)h\frac{\ln(1+h)}{h}. The values approach 1 from both sides of zero, confirming the limit.
This limit is important in many mathematical and applied contexts, including in calculus for defining derivatives, in series expansions such as the Taylor series for logarithms, and in various scientific and engineering problems where logarithmic growth or decay near a specific point is considered.
If you’d like, I can also help you see this graphically or with a more detailed explanation!